Question

Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and pure B are 55 and 15 kPa respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?

• A. 0.340
• B. 0.663
• C. 0.480
• D. 0.5217

Hint:

For ideal solutions, always combine Raoult’s law with Dalton’s law to relate liquid and vapour compositions.

Answer 1

The Correct Option is D

To find the mole fraction of liquid A in the solution (\(X_A\)) in equilibrium with a vapour where the mole fraction of A is 0.8, we can use Raoult's Law. Here's the step-by-step solution:

1. The given vapour pressures of pure liquids at temperature T K are:
   • Pure A: \(P^0_A = 55 \text{ kPa}\)
   • Pure B: \(P^0_B = 15 \text{ kPa}\)
2. We'll apply Raoult's law, which states that the partial pressure of each component in the solution is equal to the product of the mole fraction of the component in the liquid phase and the vapour pressure of the pure component. Thus, we have:
   • For component A: \(P_A = X_A \times P^0_A\)
   • For component B: \(P_B = X_B \times P^0_B\)
3. In the vapour phase, the mole fraction of A is given as 0.8, hence the mole fractions in vapour are:
   • \(Y_A = 0.8\)
   • \(Y_B = 0.2\), since \(Y_B = 1 - Y_A\)
4. According to Dalton's Law of Partial Pressures, total pressure (P) of the system can be written as: \(P = P_A + P_B\)
5. The relationship between the vapour phase and the liquid phase mole fractions and partial pressures is given by:
   • \(Y_A = \frac{P_A}{P}\)
   • \(Y_B = \frac{P_B}{P}\)
6. Now, we can express \(Y_A\) using \(P\): \(Y_A = \frac{X_A \times P^0_A}{P_A + P_B}\)
7. Substitute the known values and simplify: \(0.8 = \frac{X_A \times 55}{X_A \times 55 + (1 - X_A) \times 15}\)
8. Cross-multiply to solve for \(X_A\): \(0.8 \times (X_A \times 55 + (1 - X_A) \times 15) = X_A \times 55\)
9. On simplifying, we get: \(0.8 \times (55X_A + 15 - 15X_A) = 55X_A\)
10. Further simplification: \(44X_A + 12 = 55X_A\)
11. Rearranging the equation gives: \(11X_A = 12\)
12. Thus, solving for \(X_A\), we get: \(X_A = \frac{12}{11} = 0.5217\)

Therefore, the mole fraction of A in the solution is 0.5217, which matches the correct answer.